Question

(a) State whether or not the equation is autonomous. (b) Identify all equilibrium solutions (if any). (c) Sketch the direction field for the differential equation in the rectangular portion of the \(t y\)-plane defined by \(-2 \leq t \leq 2,-2 \leq y \leq 2\). $$ y^{\prime}=\sin y $$

Short answer

Based on the given step-by-step solution, a short answer would be: (a) The differential equation \(y' = \sin y\) is autonomous as it depends only on the variable \(y\) and not on \(t\). (b) The equilibrium solutions are given by \(y = k\pi, \quad k \in \mathbb{Z}\), as these are the values of \(y\) where the sine function equals zero. (c) To sketch the direction field, create a grid of points on the \(ty\)-plane within the specified rectangular portion, calculate the slopes at each grid point using the autonomous differential equation, and draw small line segments with the respective slopes at each point. This will help visualize the direction field for the given differential equation in the specified region.

Step-by-step solution

(a) Determine if the Equation is Autonomous

To determine if the equation is autonomous, check if the differential equation is dependent only on the variable \(y\) and not \(t\). In this case, our given differential equation is: $$ y^{\prime}=\sin y $$ This differential equation only depends on \(y\), so it is autonomous.

(b) Identify Equilibrium Solutions

Equilibrium solutions occur when the derivative is equal to zero. We can find the equilibrium solutions by setting the equation equal to zero and solving for \(y\): $$ 0=\sin y $$ The sine function equals zero at \(y = k\pi\) for all integer values of \(k\). Therefore, the equilibrium solutions are: $$ y=k\pi, \quad k \in \mathbb{Z} $$

(c) Sketch the Direction Field

To sketch the direction field for the given differential equation in the rectangular portion \(-2\leq t\leq 2\) and \(-2\leq y\leq 2\), we can follow these steps: 1. In the region, set up a grid of points on the \(ty\)-plane with a small spacing (\(\Delta t\) and \(\Delta y\)) between them. 2. At each grid point \((t,y)\), find the slope \(y^{\prime}\) using the given autonomous differential equation \(y^{\prime}=\sin y\). 3. Draw a short line segment (tangent) at each grid point with the calculated slope. For example, choose a point \((t_1, y_1)\) in the given region. Calculate the slope at that point using the equation \(y^{\prime} = \sin y_1\). Draw a small line segment with that slope at the point \((t_1, y_1)\). Finally, combine all the small line segments at different grid points to visualize the direction field for the given differential equation in the specified rectangular region.

Key concepts

Equilibrium Solutions

In the world of differential equations, equilibrium solutions are critical points that represent the state where the system remains steady over time. For autonomous differential equations like the one given by \( y^{\textprime} = \text{sin y} \), finding these solutions is a matter of setting the derivative equal to zero and solving for \( y \). This is because at equilibrium, the rate of change (derivative) of the system is zero, indicating no further movement away from that point.

Thus, when we solve \( 0 = \text{sin y} \), we find that equilibrium solutions occur at points where \( y = k\textpi \) for any integer \( k \). This means our system has an infinite number of equilibrium solutions, one at each multiple of \( \textpi \), implying that the behavior of the system repeats periodically along the \( y \)-axis. Understanding these solutions is crucial as they are often the first step in analyzing the stability and behavior of the system around these points.

Direction Field

A direction field is a graphical representation that gives us a qualitative view of the solutions of a differential equation without requiring us to solve the equation explicitly. For our autonomous equation, drawing a direction field involves plotting small line segments, which indicate the slope or direction of the solution curve at various points on the \( ty \)-plane.

To sketch this for \( y^{\textprime} = \text{sin y} \), you place a grid on the specified region and calculate the slope using the differential equation at each grid point. By doing this, you illustrate how the solution changes direction at different points. Since our equation is independent of \( t \), the line segments at a given \( y \)-value will have the same slope regardless of \( t \). This simplifies the process significantly because we only need to consider the \( y \)-values. After plotting these segments across the grid, they collectively form a pattern that visually represents all possible solution curves for the differential equation in the region of interest.

Differential Equation Sketching

Sketching a differential equation, such as \( y^{\textprime} = \text{sin y} \), involves taking both the equilibrium solutions and the direction field we have identified and combining them to visualize the possible behaviors of the system. Each small line segment in the direction field represents the instantaneous rate of change of the solution at that point. Connecting these segments we can infer the general shape of the solution curves.

When it comes to sketching, remember that the equilibrium solutions act as fixed points where the solution curve is horizontal. For our exercise, the solution curves will appear to flatten out and run parallel to the \( t \)-axis at each \( y = k\textpi \) point. Meanwhile, in other regions, the direction and curvature of the solution curves can indicate whether the solutions are moving towards or away from these equilibrium points. This visual aid is invaluable for understanding the system's dynamics without solving the equation analytically and can provide insights into the stability and long-term behavior of solutions.